A) 250
B) 400
C) 300
D) 200
Correct Answer: D
Solution :
[d] As we know, \[\left[ \frac{N}{{{N}_{0}}} \right]={{\left[ \frac{1}{2} \right]}^{n}}\] ?(i) n = no. of half life N - no. of atoms left \[{{N}_{0}}\] - initial no. of atoms By radioactive decay law, \[\frac{dN}{dt}=kN\] k - disintegration constant \[\therefore \]\[\frac{\frac{dN}{dt}}{\frac{d{{N}_{0}}}{dt}}=\frac{N}{{{N}_{0}}}\] ?(ii) From (i) and (ii) we get \[\therefore \]\[\frac{\frac{dN}{dt}}{\frac{d{{N}_{0}}}{dt}}={{\left[ \frac{1}{2} \right]}^{n}}\] or, \[\left[ \frac{100}{1600} \right]={{\left[ \frac{1}{2} \right]}^{n}}\] \[\Rightarrow \]\[{{\left[ \frac{1}{2} \right]}^{4}}={{\left[ \frac{1}{2} \right]}^{n}}\] \[\therefore \]\[n=4,\] Therefore, in 8 seconds 4 half-life had occurred in which counting rate reduces to 100 counts\[{{s}^{-1}}\]. \[\therefore \] Half life, \[{{T}_{\frac{1}{2}}}=2\,\sec \] In 6 sec, 3 half-life will occur \[\left[ \frac{\frac{dN}{dt}}{1600} \right]={{\left[ \frac{1}{2} \right]}^{3}}\] \[\Rightarrow \]\[\frac{dN}{dt}=200\,\,\text{count}{{\text{s}}^{-1}}\]
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